; Exercise 1.37. a. An infinite continued fraction is an expression of the form

;

;

;

; As an example, one can show that the infinite continued fraction expansion with the Ni and the Di; all equal to 1 produces 1/phi , where phi is the golden ratio (described in section 1.2.2). One way

; to approximate an infinite continued fraction is to truncate the expansion after a given number of

; terms. Such a truncation -- a so-called k-term finite continued fraction -- has the form

;

;

; Suppose that n and d are procedures of one argument (the term index i) that return the Ni and Di of the; terms of the continued fraction. Define a procedure cont-frac such that evaluating (cont-frac n d k)

; computes the value of the k-term finite continued fraction. Check your procedure by approximating

; 1/phi using

;

; (cont-frac (lambda (i) 1.0)

; (lambda (i) 1.0)

; k)

;

; for successive values of k. How large must you make k in order to get an approximation that is accurate

; to 4 decimal places?

;

; b. If your cont-frac procedure generates a recursive process, write one that generates an iterative

; process. If it generates an iterative process, write one that generates a recursive process.

(define (cont-frac-recur n d k)

(define (iter i)

(if (> i k)

0

(/ (n i) (+ (d i) (iter (+ i 1))))))

(iter 1))

(display (cont-frac-recur (lambda(i) 1.0) (lambda(i) 1.0) 100))(newline)

(define (cont-frac-iter n d k)

(define (iter i result)

(if (= i 0)

result

(iter (- i 1) (/ (n i) (+ (d i) result)))))

(iter k 0))

(display (cont-frac-iter (lambda(i) 1.0) (lambda(i) 1.0) 1000))(newline)